Civil Engineering 2024 Paper I 50 marks Solve

Q2

(a) A cantilever beam ABC as shown in the above figure is having a total span of 2·0 m. The maximum safe allowable bending stress is 7500 kN/m² for the material. Find the maximum safe uniformly distributed load which this beam can carry. What will be the maximum shear stress at support A for the obtained safe UDL ? (Neglect the self weight of beam) (20 marks) (b) From first principles, derive the expression for determining the depth of neutral axis, for a rectangular reinforced concrete section without compression reinforcement, as per Limit State Method. Use the stress-strain curves for concrete and reinforcing bars shown in the Figs. 1 and 2. (10 marks) (c) Two angles ISA 100×100×12 mm transmit an ultimate tensile force of 540 kN, acting through the C.G. of angle sections as shown in the Figure. The angles are connected to the gusset plate on either side by welding. Design the lengths l₁ and l₂ of the weld if the size of the fillet weld is 6 mm, fᵤ = 410 MPa, partial safety factor for the weld γₘw = 1·25. Relevant portion of the IS 800 : 2007 is enclosed. (20 marks)

हिंदी में प्रश्न पढ़ें

(a) नीचे चित्र में दर्शाई गई प्रास धरन ABC की कुल विस्तृति 2 मीटर है । पदार्थ के लिए अधिकतम सुरक्षित अनुजेय बंकन प्रतिबल 7500 kN/m² है । उस अधिकतम सुरक्षित एक समान वितरित भार को ज्ञात कीजिए जिसे यह धरन वहन कर सकती है । प्राप्त किए गए सुरक्षित एकसमान वितरित भार के लिए आलम्ब A पर अधिकतम अपरूपण प्रतिबल कितना होगा ? (धरन का स्वभार नगण्य है) (20 अंक) (b) सीमान्त-अवस्था-अभिकल्पन विधि द्वारा, बिना संपीडन प्रबलन वाले एक आयताकार प्रबलित कंक्रीट काट के लिए, प्रथम सिद्धान्त के द्वारा, उदासीन अक्ष की गहराई के निर्धारण के लिए व्यंजक व्युत्पन्न कीजिए । चित्र 1 एवं 2 में क्रमशः: कंक्रीट एवं प्रबलन छड़ों के लिए दर्शाए गए वक्रों का उपयोग कीजिए । (10 अंक) (c) दो लोह कोण ISA 100×100×12 mm, चित्र में दर्शाए अनुसार, लोह कोण के गुरुत्व केन्द्र पर लगने वाले 540 kN के चरम तनन बल को प्रेषित करते हैं। लोह कोण, संगम पट्टिका के दोनों ओर वेल्डिंग द्वारा जोड़े गए हैं। यदि फिलेट वेल्ड का आमाप 6 mm हो तो लम्बाई l₁ एवं l₂ की अभिकल्पना कीजिए। fᵤ = 410 MPa, वेल्डिंग के लिए आंशिक सुरक्षा गुणक γₘw = 1·25। IS 800 : 2007 का संबंधित भाग संलग्न है। (20 अंक)

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Approach

Begin with a clear statement of given data for each sub-part. For part (a), apply bending stress formula σ = My/I and shear stress formula τ = VQ/(Ib) systematically. For part (b), derive the neutral axis depth using strain compatibility and equilibrium of forces from first principles with proper stress-strain diagrams. For part (c), design weld lengths using IS 800:2007 provisions for eccentrically loaded welded connections. Allocate approximately 40% time to part (a), 25% to part (b), and 35% to part (c) based on marks distribution.

Key points expected

  • Part (a): Correct identification of maximum bending moment location and value for cantilever under UDL; calculation of section modulus; determination of safe UDL using σ = M/Z; computation of maximum shear stress at support using τ = VQ/(Ib) or simplified formula
  • Part (a): Proper handling of beam geometry (ABC with possible overhang or varying section) and correct interpretation of 'maximum safe allowable bending stress' as working stress
  • Part (b): Derivation starting from strain distribution (linear, εc at top, εs at steel level), stress-strain relationship for concrete (parabolic-rectangular or as per given Fig. 1), stress-strain for steel (bilinear or as per given Fig. 2)
  • Part (b): Force equilibrium equation: Cc = T → 0.36fckbxu = 0.87fyAst; solving for neutral axis depth xu = (0.87fyAst)/(0.36fckb); clear definition of all terms
  • Part (c): Identification of weld configuration (two angles, welds on both sides of gusset); calculation of throat thickness = 0.7×size; determination of weld strength per unit length = fu/(√3×γmw); resolution of force components and moments about C.G.
  • Part (c): Application of IS 800:2007 clause on eccentrically loaded welds: calculation of direct shear and torsional shear, vectorial combination, and iterative or direct solution for lengths l₁ and l₂ satisfying strength requirements

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly identifies that part (a) requires working stress method (σ = M/Z) while part (b) uses limit state method; recognizes part (c) involves eccentrically loaded fillet welds per IS 800:2007; distinguishes between allowable stress design and limit state philosophy; correctly interprets 'safe' in part (a) as working stress approachMixes up working stress and limit state concepts; uses partial safety factors in part (a) or omits them where required; basic understanding of weld design but confuses direct and torsional shear componentsApplies limit state method to part (a) or working stress to part (b); fundamental confusion about neutral axis derivation; treats weld as concentrically loaded or ignores eccentricity completely
Numerical accuracy20%10Part (a): Correct safe UDL calculation with proper units (kN/m) and maximum shear stress at support; Part (b): Accurate algebraic derivation leading to xu = (0.87fyAst)/(0.36fckb); Part (c): Precise weld strength calculation (410/(√3×1.25) = 189.37 MPa), correct throat thickness (4.2 mm), and accurate length determination for l₁ and l₂Minor arithmetic errors in one part; correct method but wrong final values; unit conversion errors (kN/m² to MPa); approximate values for weld strength without showing calculationMajor calculation errors in multiple parts; wrong formulas applied; incorrect order of magnitude in answers; missing critical steps like throat thickness or partial safety factor application
Diagram quality20%10Part (a): Clear cantilever beam diagram with loading, reactions, and shear force/bending moment diagrams; Part (b): Accurate stress-strain curves for concrete (parabolic-rectangular) and steel (bilinear with yield plateau), strain diagram, stress block diagram, and force equilibrium diagram; Part (c): Weld configuration showing angles, gusset plate, weld lengths l₁ and l₂, and force eccentricityBasic diagrams present but missing labels or dimensions; incomplete stress-strain curves; weld diagram without showing eccentricity or C.G. locationMissing critical diagrams especially for part (b); no free body diagrams; diagrams without any labels; hand-drawn sketches that are illegible or conceptually wrong
Step-by-step derivation20%10Part (a): Clear progression from loading → reactions → SFD → BMD → section modulus → stress check; Part (b): Systematic derivation: assumptions → strain compatibility → stress distribution → force equilibrium → solution for xu with all intermediate steps; Part (c): Methodical application of IS 800:2007: weld strength → load decomposition → eccentricity calculation → combined stress check → length determinationSome steps skipped or combined; derivation present but missing justification for key assumptions; final answer correct but intermediate steps unclearNo derivation shown for part (b)—only final formula stated; jumps directly to answers without showing working; missing critical steps like equilibrium equations or compatibility conditions
Practical interpretation20%10Part (a): Comments on shear stress distribution and why maximum occurs at neutral axis; Part (b): Discusses implications of xu < xu,max for under-reinforced section and ductile failure; Part (c): Checks weld lengths against practical minimums (IS 800:2007 clause 10.5.4.2), discusses load path through gusset plate, and notes importance of balancing weld lengths for uniform force distributionBrief comment on one part; mentions code provisions without elaboration; basic awareness of practical constraintsNo practical interpretation; purely mathematical treatment; fails to recognize that derived weld lengths must satisfy minimum length requirements; no discussion of failure modes

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